# Program to find frequency of each element in a vector using map in C++

Given a vector **vec**, the task is to find the frequency of each element of **vec** using a map.

**Examples: **

Input: vec = {1, 2, 2, 3, 1, 4, 4, 5}

Output:

1 2

2 2

3 1

4 2

5 1

Explanation:

1 has occured 2 times

2 has occured 2 times

3 has occured 1 times

4 has occured 2 times

5 has occured 1 times

Input: v1 = {6, 7, 8, 6, 4, 1}

Output:

1 1

4 1

6 2

7 1

8 1

Explanation:

1 has occured 1 times

4 has occured 1 times

6 has occured 2 times

7 has occured 1 times

8 has occured 1 times

**Approach: **

We can find the frequency of elements in a vector using given four steps efficiently:

- Traverse the elements of the given vector
**vec**. - check whether the current element is present in map or not.
- If it is present, then update the frequency of the current element, else insert the element with frequency 1 as shown below:
- Traverse the map and print the frequency of each element stored as a mapped value.

if(M.find(vec[i])==M.end()) { M[vec[i]]=1; } else { M[vec[i]]++; }

Below is the implementation of the above approach:

## CPP

`#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `void` `printFrequency(vector<` `int` `> vec) ` `{ ` ` ` `// Define an map ` ` ` `map<` `int` `, ` `int` `> M; ` ` ` ` ` `// Traverse vector vec check if ` ` ` `// current element is present ` ` ` `// or not ` ` ` `for` `(` `int` `i = 0; vec[i]; i++) { ` ` ` ` ` `// If the current element ` ` ` `// is not found then insert ` ` ` `// current element with ` ` ` `// frequency 1 ` ` ` `if` `(M.find(vec[i]) == M.end()) { ` ` ` `M[vec[i]] = 1; ` ` ` `} ` ` ` ` ` `// Else update the frequency ` ` ` `else` `{ ` ` ` `M[vec[i]]++; ` ` ` `} ` ` ` `} ` ` ` ` ` `// Traverse the map to print the ` ` ` `// frequency ` ` ` `for` `(` `auto` `& it : M) { ` ` ` `cout << it.first << ` `' '` ` ` `<< it.second << ` `'\n'` `; ` ` ` `} ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `vector<` `int` `> vec = { 1, 2, 2, 3, 1, 4, 4, 5 }; ` ` ` ` ` `// Function call ` ` ` `printFrequency(vec); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

1 2 2 2 3 1 4 2 5 1

**Complexity Analysis:**

**Time Complexity:** O(n)

For a given vector of size n, we are iterating over it once. So the time complexity is O(n)

**Space Complexity:** O(n)

For a given vector of size n, we are using an extra map which can have maximum of n key-values, so space complexity is O(n)

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